Arithmetic series/progression problems Watch

lolface12
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Did new topic in class today which i found so so hard. I have been given some questions to do but no idea how to do it.


Can anybody please help me?

No. 1:


"A bank account is opened at the beginning of the year 2013 by making a deposit of £1000. Further deposits of £500 are made at the beginning of the subsequent year. The rate of interest is 2.5% and the interest is paid at the end of each year. Find the total amount gathered in the account at the end of year 2030, just after the interest is paid"

How to solve this? it says it is 15 marks so there must be lots of calculations. i dont know how to do it. anyone can please help?

No. 2:

Find the minimum no. of terms of the geometric sequence 1.5, 2.25, 3.375, ... which will make the terms more than 1000? (6)

Any idea of how to do this aswell? I am so stuck!
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brianeverit
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(Original post by lolface12)
Did new topic in class today which i found so so hard. I have been given some questions to do but no idea how to do it.


Can anybody please help me?

No. 1:


"A bank account is opened at the beginning of the year 2013 by making a deposit of £1000. Further deposits of £500 are made at the beginning of the subsequent year. The rate of interest is 2.5% and the interest is paid at the end of each year. Find the total amount gathered in the account at the end of year 2013, just after the interest is paid"

How to solve this? it says it is 15 marks so there must be lots of calculations. i dont know how to do it. anyone can please help?

No. 2:

Find the minimum no. of terms of the geometric sequence 1.5, 2.75, 3.375, ... which will make the terms more than 1000? (6)

Any idea of how to do this aswell? I am so stuck!
Are you sure you have copied the question correctly. If you have then the adding of £500 at th beginning of each year does not make sense because there are no further years.If the latter date should have been later than 2013 t5rhe proceed as follows

I would start by listing it year by year
i.e.
Amount at end of year 1==?
Interest added=?
Amount at start of year 2=?
Amount at end of year 2=?
Interest added=?
Amount at start of year 3=? etc and see if you can see any patterns.
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lolface12
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(Original post by brianeverit)
Are you sure you have copied the question correctly. If you have then the adding of £500 at th beginning of each year does not make sense because there are no further years.If the latter date should have been later than 2013 t5rhe proceed as follows

I would start by listing it year by year
i.e.
Amount at end of year 1==?
Interest added=?
Amount at start of year 2=?
Amount at end of year 2=?
Interest added=?
Amount at start of year 3=? etc and see if you can see any patterns.
I am sorry. The year is 2030. How to tackle this question please?
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lolface12
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Please can someone explain to me how to do this? 15 marks is alot and i dont know what answers and working out is neeeded
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ztibor
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(Original post by lolface12)
Please can someone explain to me how to do this? 15 marks is alot and i dont know what answers and working out is neeeded
Beg of 2013:1000
End of 2013: 1000*(1-0.025)=1000*0.975 (1st year end)
Beg of 2014: 1000*0.975+500
End of 2014: (1000*0.975+500)*0.975=1000*0.97 5^2 + 500*0.975 (2nd year end)
beg of 2015: 1000*0.975^2 + 500*0.975+500
End of 2014: (1000*0.975^2 + 500*0.975+500)*0.975=1000*0.975^ 3+500(0.975+0975^2)
........
......
End of the n th year :
\displaystyle 1000 \cdot 0.975^n +500\cdot (0.975 +0.975^2 + ... + 0.975^{n-1})=
\displaystyle =1000 \cdot 0.975^n +500\cdot 0.975 \cdot (1 +0.975 + ... + 0.975^{n-2})

Use the formula for sum of first (n-1) terms of geometric sequence. When q is the quotient
and a_1 is the first term, then
\displaystyle s_{n-1}=a_1\cdot \frac{q^{n-1}-1}{q-1}
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lolface12
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Can somebody please explain where i am going wrong with question 2?
What do i do from this point?




1st term is 1.5 and common ratio is also 1.5 i think. What do i do next? Did i make mistake?
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lolface12
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I saw this example on google to find out where i am going wrong




I don't know how he got from the 1st to the 2nd line, or from 2nd to 3rd.
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lolface12
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(Original post by ztibor)
Beg of 2013:1000
End of 2013: 1000*(1-0.025)=1000*0.975 (1st year end)
Beg of 2014: 1000*0.975+500
End of 2014: (1000*0.975+500)*0.975=1000*0.97 5^2 + 500*0.975 (2nd year end)
beg of 2015: 1000*0.975^2 + 500*0.975+500
End of 2014: (1000*0.975^2 + 500*0.975+500)*0.975=1000*0.975^ 3+500(0.975+0975^2)
........
......
End of the n th year :
\displaystyle 1000 \cdot 0.975^n +500\cdot (0.975 +0.975^2 + ... + 0.975^{n-1})=
\displaystyle =1000 \cdot 0.975^n +500\cdot 0.975 \cdot (1 +0.975 + ... + 0.975^{n-2})

Use the formula for sum of first (n-1) terms of geometric sequence. When q is the quotient
and a_1 is the first term, then
\displaystyle s_{n-1}=a_1\cdot \frac{q^{n-1}-1}{q-1}
So what are the figures for Sn, Q and n. I know a is 1000. Still confused about how to work out total amount gathered
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ztibor
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(Original post by lolface12)
So what are the figures for Sn, Q and n. I know a is 1000. Still confused about how to work out total amount gathered
What do you think to be s_n . I wrote s_n as sum of a geometric sequence
The cummulated amount in the bank account is not s_n.
To calculate it i wrote a formula
First part of this formula is a power, the second (in the brackets) is a sum of geometric sequence,
multiplied by some constants, which sum I wrote in a separate formula bellow that.
(and that summarize only n-1 terms in the sequence as you need for solution)
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lolface12
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(Original post by lolface12)
Can somebody please explain where i am going wrong with question 2?
What do i do from this point?




1st term is 1.5 and common ratio is also 1.5 i think. What do i do next? Did i make mistake?
(Original post by lolface12)
I saw this example on google to find out where i am going wrong




I don't know how he got from the 1st to the 2nd line, or from 2nd to 3rd.

Please can someone explain me how to get answer here and where i am going wrong? It is for the 2nd question in my post.
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lolface12
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Anybody at all?
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cambo211
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(Original post by lolface12)
I saw this example on google to find out where i am going wrong




I don't know how he got from the 1st to the 2nd line, or from 2nd to 3rd.
1st to 2nd he just neatened it up.

2nd to 3rd he multiplied the LHS by (-10/-10) to get rid of the fraction.
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cambo211
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(Original post by lolface12)
Can somebody please explain where i am going wrong with question 2?
What do i do from this point?




1st term is 1.5 and common ratio is also 1.5 i think. What do i do next? Did i make mistake?
 1.5^n\times1.5\not=2.25^n
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lolface12
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(Original post by cambo211)
 1.5^n\times1.5\not=2.25^n
Is that 1.5 x 1.5 the next line and if so how does that come about? Do i do 2.25n divided by 1.5n too?

What do i do with the 1000 on the RHS? Cheers mate
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cambo211
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(Original post by lolface12)
Is that 1.5 x 1.5 the next line and if so how does that come about? Do i do 2.25n divided by 1.5n too?

What do i do with the 1000 on the RHS? Cheers mate
Sorry i should have made myself clearer.

You wrote that  1.5^n\times1.5=2.25^n. This is a mistake.

 3^2\times3\not=9^2

3^2\times3=3^3

See where you've gone wrong with that bit?
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lolface12
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(Original post by cambo211)
Sorry i should have made myself clearer.

You wrote that  1.5^n\times1.5=2.25^n. This is a mistake.

 3^2\times3\not=9^2

3^2\times3=3^3

See where you've gone wrong with that bit?
Yes i think so mate, but not sure what the answer should be if it's not -2.25. I am struggling to see it, and what to do next after with the 1000 on the other side
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cambo211
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(Original post by lolface12)
Yes i think so mate, but not sure what the answer should be if it's not -2.25. I am struggling to see it, and what to do next after with the 1000 on the other side
3^2\times3=3^{2+1}

Any clearer on where we should be?

Need to get it in a form we can proceed with before you even need to worry about the 1000.

So long as what we're doing to the LHS doesn't actually change it the 1000 stays as it is.
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lolface12
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(Original post by cambo211)
3^2\times3=3^{2+1}

Any clearer on where we should be?

Need to get it in a form we can proceed with before you even need to worry about the 1000.

So long as what we're doing to the LHS doesn't actually change it the 1000 stays as it is.

Would it be 1.5^{n+1}
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cambo211
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(Original post by lolface12)
Would it be1.5^{n+1}
Yep. 1.5^{n+1}.

So now we have  \dfrac{1.5-1.5^{n+1}}{-0.5}>1000

We can now get rid of the fraction.
For the sake of keeping the inequality simple we can multiply the LHS by \dfrac{-1}{-1} to bring us to a point where we can easily move forward. Can you see why this doesn't actually change the value of the LHS at all?

 \dfrac{1.5^{n+1}-1.5}{0.5}>1000

Any idea where to go next?
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lolface12
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(Original post by cambo211)
Yep. 1.5^{n+1}.

So now we have  \dfrac{1.5-1.5^{n+1}}{-0.5}>1000

We can now get rid of the fraction.
For the sake of keeping the inequality simple we can multiply the LHS by \dfrac{-1}{-1} to bring us to a point where we can easily move forward.

Any idea where to go next?
Is multiplying the LHS here done to get rid of the fraction? I know that near the end i would Ln to work out what n is and get the answer from there, i know too simplify 1000 before that too. But it's just the getting there part which i am struggling to grasp. Thanks for your help so far.

In the example i am struggling to work out how he got rid of the fraction by multiplying it by -10/-10 in the LHS. It's probably really simple but i just can't see it and what to do from that point
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